题目内容
请比较1+
+
+
+…+
与
+
+
+
+…+
的大小.
1 |
2 |
2 |
3 |
3 |
4 |
2007 |
2008 |
2 |
1 |
3 |
2 |
4 |
3 |
5 |
4 |
1984 |
1983 |
考点:分数的大小比较
专题:计算问题(巧算速算)
分析:先把算式利用
=1-
化简为:1+
+
+
+…+
=1+1-
+1-
+…+1-
=2008-(
+
+…+
)
+
+
+
+…+
=1+1+1+
+1+
+…1+
=1984+
+
+…
;
所以:
1+
+
+
+…+
-(
+
+
+
+…+
)
=2008-1984-(
+
+…
+
+
+…
)
然后利用:“欧拉公式”:
+
+…
=ln(2008)+C-1=7.1821,(其中C为欧拉常数 数值是0.5772…).
+
+…
=ln(1983)+C-1=7.1696,看这个差如果大于0,第一个算式结果答,如果等于0,这两个算式结果相同,差比0小,说明后面的算式结果大.
n |
n+1 |
1 |
n+1 |
1 |
2 |
2 |
3 |
3 |
4 |
2007 |
2008 |
1 |
2 |
1 |
3 |
1 |
2008 |
1 |
2 |
1 |
3 |
1 |
2008 |
2 |
1 |
3 |
2 |
4 |
3 |
5 |
4 |
1984 |
1983 |
1 |
2 |
1 |
3 |
1 |
1983 |
1 |
2 |
1 |
3 |
1 |
1983 |
所以:
1+
1 |
2 |
2 |
3 |
3 |
4 |
2007 |
2008 |
2 |
1 |
3 |
2 |
4 |
3 |
5 |
4 |
1984 |
1983 |
=2008-1984-(
1 |
2 |
1 |
3 |
1 |
2008 |
1 |
2 |
1 |
3 |
1 |
1983 |
然后利用:“欧拉公式”:
1 |
2 |
1 |
3 |
1 |
2008 |
1 |
2 |
1 |
3 |
1 |
1983 |
解答:
解:1+
+
+
+…+
=1+1-
+1-
+…+1-
=2008-(
+
+…+
)
+
+
+
+…+
=1+1+1+
+1+
+…1+
=1984+
+
+…
所以:
1+
+
+
+…+
-(
+
+
+
+…+
)
=2008-1984-(
+
+…
+
+
+…
)
=24-(8.1821-1+8.1696-1)
=24-14.3517
=9.6483>0
答:1+
+
+
+…+
>
+
+
+
+…+
1 |
2 |
2 |
3 |
3 |
4 |
2007 |
2008 |
=1+1-
1 |
2 |
1 |
3 |
1 |
2008 |
=2008-(
1 |
2 |
1 |
3 |
1 |
2008 |
2 |
1 |
3 |
2 |
4 |
3 |
5 |
4 |
1984 |
1983 |
=1+1+1+
1 |
2 |
1 |
3 |
1 |
1983 |
=1984+
1 |
2 |
1 |
3 |
1 |
1983 |
所以:
1+
1 |
2 |
2 |
3 |
3 |
4 |
2007 |
2008 |
2 |
1 |
3 |
2 |
4 |
3 |
5 |
4 |
1984 |
1983 |
=2008-1984-(
1 |
2 |
1 |
3 |
1 |
2008 |
1 |
2 |
1 |
3 |
1 |
1983 |
=24-(8.1821-1+8.1696-1)
=24-14.3517
=9.6483>0
答:1+
1 |
2 |
2 |
3 |
3 |
4 |
2007 |
2008 |
2 |
1 |
3 |
2 |
4 |
3 |
5 |
4 |
1984 |
1983 |
点评:本题考查分数运算的巧算,注意根据分数的特点去巧妙计算即可.
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